It is asked in the posting to answer (ii) and (iii) . To answer (iii) Figure is not visible , hence we are unable to answer (iii).
So only (ii) is answered below.
3 μF capacitor is connected to 90V battery through 200Ω resistance.
Hence charge acquired by the capacitor will be 3×90 = 270 μC.
6 μF capacitor is connected across 100Ω resistance. Once the 3 μF capacitor is charged there will not be any current flow
through this 100Ω resistor, hence ther will not be any potential drop across this 100Ω resistor.
Hence 6 μF capacitor will not be charged